(e^( cos^(-1) (log x^2) ))^1/2
The base of log is 4
$\displaystyle f(x) = \sqrt{e^{\arccos{[\log{(x^2)}]}}}$.
Note that the logarithm is only defined for $\displaystyle x > 0$.
Since $\displaystyle x^2 \geq 0$, we can already see that we can not let $\displaystyle x = 0$.
Also note that $\displaystyle \arccos{x}$ is only defined for $\displaystyle -1 \leq x \leq 1$.
The exponential function is always $\displaystyle >0$, so there are not any further restrictions for the square root.
Thus the domain is $\displaystyle x \in [-1, 0)\cup(0, 1]$.